0.1 Pre-Calculus Review
The algebra, function and trigonometry toolkit that every page of calculus leans on — with labs to make each idea visible.
Calculus is built on a small set of algebra, function and trigonometry skills. Every derivative rule, limit trick and integration technique in this course reuses them. This review collects exactly what you need — no more — with interactive labs so you can see each idea. Skim what you know, slow down where you don't, and try the exercises at the end.
🧭 How to use this page
- Took the diagnostic on the Chapter 0 introduction? Jump straight to the topics it suggested.
- Hover (or tap) any dotted word for its Arabic name and a plain-English meaning; press ع (bottom-right) to show Arabic beside every term.
- Topics: 1 Numbers & intervals · 2 Functions · 3 Graphs & transformations · 4 Combining functions · 5 Lines & quadratics · 6 Factoring & fractions · 7 Exponents & logarithms · 8 Inverse functions · 9 Trigonometry · Exercises
1. Real Numbers and Intervals
We work with the real numbers $\mathbb R$: every point on the number line. Sets of real numbers between two end points are written in interval notation. A round bracket means the end point is excluded; a square bracket means it is included. Infinity always gets a round bracket, because $\infty$ is not a number you can reach.
| Interval | Inequality | Picture |
|---|---|---|
| $(a,b)$ | $a<x<b$ | open at both ends ○—○ |
| $[a,b]$ | $a\le x\le b$ | closed at both ends ●—● |
| $[a,b)$ | $a\le x<b$ | ●—○ |
| $(a,\infty)$ | $x>a$ | ○———→ |
| $(-\infty,b]$ | $x\le b$ | ←———● |
| $(-\infty,\infty)$ | all real $x$ | ←———→ |
The absolute value $|x|$ is the distance from $x$ to $0$: $|x|=x$ if $x\ge0$ and $|x|=-x$ if $x<0$. So $|x-a|<\delta$ means “$x$ is within $\delta$ of $a$”, i.e. $a-\delta<x<a+\delta$ — the language of limits.
2. Functions, Domain and Range
📘 Definition — Function
A function $f$ is a rule that assigns to each input $x$ in a set $D$ exactly one output $f(x)$. The set $D$ is the domain; the set of all outputs is the range. On a graph: a curve is the graph of a function exactly when every vertical line meets it at most once (the vertical line test).
🧭 Finding the natural domain
Start with all real numbers and remove the values that break a rule:
- Division: the denominator cannot be $0$.
- Even roots: $\sqrt{\;\cdot\;}$ needs the inside $\ge0$.
- Logarithms: $\ln(\;\cdot\;)$ needs the inside $>0$.
- $\tan x$ and $\sec x$ exclude $x=\tfrac\pi2+k\pi$; $\cot x$ and $\csc x$ exclude $x=k\pi$.
✏️ Example 1 — Domain and range
Find the domain and range of $f(x)=\sqrt{x-2}$ and the domain of $g(x)=\dfrac{1}{x^2-4}$.
💡 Show solution▶
$f$ Need $x-2\ge0$, so the domain is $[2,\infty)$. A square root is never negative and takes every value $\ge0$, so the range is $[0,\infty)$.
$g$ Need $x^2-4\ne0$, i.e. $x\ne\pm2$. Domain: $(-\infty,-2)\cup(-2,2)\cup(2,\infty)$.
✏️ Example 2 — Evaluating with expressions — the heart of the derivative
For $f(x)=x^2-3x$, find and simplify $\dfrac{f(a+h)-f(a)}{h}$, $h\ne0$.
💡 Show solution▶
Substitute $f(a+h)=(a+h)^2-3(a+h)=a^2+2ah+h^2-3a-3h$.
Subtract $f(a+h)-f(a)=2ah+h^2-3h$.
Divide $\dfrac{2ah+h^2-3h}{h}=2a+h-3$. This difference quotient is exactly what becomes the derivative in Chapter 2 (let $h\to0$: $2a-3$).
3. The Function Library and Transformations
A handful of basic graphs appear again and again. Know their shapes, domains and key points by heart: $x^2$ (parabola), $x^3$, $\sqrt x$, $\dfrac1x$, $|x|$, $e^x$, $\ln x$, $\sin x$, $\cos x$. Every other graph in this course is built from these by transformations:
| Formula | Effect on the graph of $y=f(x)$ |
|---|---|
| $f(x)+k$ | shift up $k$ (down if $k<0$) |
| $f(x-h)$ | shift right $h$ (left if $h<0$) — note the minus sign! |
| $a\,f(x)$ | vertical stretch by $|a|$; reflect in the $x$-axis if $a<0$ |
| $f(bx)$ | horizontal compression by $|b|$; reflect in the $y$-axis if $b<0$ |
4. Combining Functions: Composition
Functions can be added, subtracted, multiplied and divided pointwise: $(f+g)(x)=f(x)+g(x)$, $(fg)(x)=f(x)g(x)$, and so on. More important for calculus is composition: feed the output of one function into another, $$(f\circ g)(x)=f\big(g(x)\big).$$ First apply $g$ (the inner function), then $f$ (the outer function). In general $f\circ g\ne g\circ f$. Recognising inner and outer functions is exactly the skill the chain rule and substitution need.
✏️ Example 3 — Composing and decomposing
Let $f(x)=\sqrt x$ and $g(x)=x^2+1$. Find $(f\circ g)(x)$ and $(g\circ f)(x)$. Then write $h(x)=\left(3x-1\right)^5$ as a composition.
💡 Show solution▶
$f\circ g$ $f\big(g(x)\big)=\sqrt{x^2+1}$, defined for all $x$.
$g\circ f$ $g\big(f(x)\big)=\left(\sqrt x\right)^2+1=x+1$, but only for $x\ge0$ (the domain of the inner function $\sqrt x$).
Decompose $h=F\circ G$ with inner $G(x)=3x-1$ and outer $F(u)=u^5$.
5. Lines and Quadratics
The slope of the line through $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\dfrac{y_2-y_1}{x_2-x_1}$ (rise over run). Useful forms of a line:
Point–slope form
$$y-y_1=m(x-x_1)$$ The form we use for every tangent line: $y=f(a)+f^\prime(a)(x-a)$.
Slope–intercept form
$$y=mx+b$$ Parallel lines share $m$; perpendicular slopes satisfy $m_1m_2=-1$.
A quadratic $ax^2+bx+c$ ($a\ne0$) has a parabola as its graph, with vertex at $x=-\dfrac{b}{2a}$. Its roots come from factoring or from the quadratic formula $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ The discriminant $b^2-4ac$ tells you how many real roots: two ($>0$), one ($=0$) or none ($<0$).
✏️ Example 4 — A line and a quadratic
(a) Find the line through $(1,2)$ and $(3,8)$. (b) Solve $2x^2+3x-2=0$ and find the vertex of $y=x^2-4x+1$.
💡 Show solution▶
(a) $m=\dfrac{8-2}{3-1}=3$; $y-2=3(x-1)$, i.e. $y=3x-1$.
(b) $2x^2+3x-2=(2x-1)(x+2)=0$, so $x=\tfrac12$ or $x=-2$. For $y=x^2-4x+1$: $x=-\tfrac{-4}{2}=2$, $y=4-8+1=-3$; vertex $(2,-3)$.
6. Factoring, Fractions and Conjugates
Most $\tfrac00$ limits in Chapter 1 are resolved with three algebra moves. Keep these patterns handy:
| Pattern | Factorisation |
|---|---|
| Difference of squares | $a^2-b^2=(a-b)(a+b)$ |
| Perfect squares | $a^2\pm2ab+b^2=(a\pm b)^2$ |
| Difference / sum of cubes | $a^3-b^3=(a-b)(a^2+ab+b^2)$, $\;a^3+b^3=(a+b)(a^2-ab+b^2)$ |
| Grouping | $x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x+1)(x-2)(x+2)$ |
| Trinomial $ax^2+bx+c$ | find two numbers with product $ac$ and sum $b$ |
✏️ Example 5 — The three limit-algebra moves
Simplify (a) $\dfrac{x^2-9}{x^2+x-6}$, (b) $\dfrac{\sqrt{x+4}-2}{x}$, (c) $\dfrac{\frac1x-\frac1{x+h}}{h}$.
💡 Show solution▶
(a) Factor $\dfrac{(x-3)(x+3)}{(x+3)(x-2)}=\dfrac{x-3}{x-2}$ for $x\ne-3$.
(b) Conjugate Multiply top and bottom by $\sqrt{x+4}+2$: $\dfrac{(x+4)-4}{x\left(\sqrt{x+4}+2\right)}=\dfrac{1}{\sqrt{x+4}+2}$ for $x\ne0$.
(c) Common denominator $\dfrac1x-\dfrac1{x+h}=\dfrac{(x+h)-x}{x(x+h)}=\dfrac{h}{x(x+h)}$, so the whole expression is $\dfrac{1}{x(x+h)}$.
7. Exponents and Logarithms
Laws of exponents ($a,b>0$)
$a^xa^y=a^{x+y}$, $\quad\dfrac{a^x}{a^y}=a^{x-y}$, $\quad(a^x)^y=a^{xy}$
$(ab)^x=a^xb^x$, $\quad a^0=1$, $\quad a^{-x}=\dfrac1{a^x}$, $\quad a^{m/n}=\sqrt[n]{a^m}$
$(ab)^x=a^xb^x$, $\quad a^0=1$, $\quad a^{-x}=\dfrac1{a^x}$, $\quad a^{m/n}=\sqrt[n]{a^m}$
Laws of logarithms ($x,y>0$)
$\log_b(xy)=\log_bx+\log_by$, $\quad\log_b\dfrac xy=\log_bx-\log_by$
$\log_b(x^r)=r\log_bx$, $\quad\log_bx=\dfrac{\ln x}{\ln b}$ (change of base)
$\log_b(x^r)=r\log_bx$, $\quad\log_bx=\dfrac{\ln x}{\ln b}$ (change of base)
The logarithm undoes the exponential: $y=\log_bx\iff b^y=x$. With the special base $e\approx2.71828$ we write $\ln x=\log_ex$, so $e^{\ln x}=x$ ($x>0$) and $\ln(e^x)=x$. The graphs of $b^x$ and $\log_bx$ are mirror images in the line $y=x$ — see below.
✏️ Example 6 — Solving exponential and logarithmic equations
Solve (a) $e^{2x}=7$, (b) $\log_3x+\log_3(x-8)=2$.
💡 Show solution▶
(a) Take $\ln$ of both sides: $2x=\ln7$, so $x=\dfrac{\ln7}{2}\approx0.973$.
(b) Combine: $\log_3\big(x(x-8)\big)=2$, so $x^2-8x=9$, $x^2-8x-9=(x-9)(x+1)=0$. Since $\log_3x$ needs $x>0$ (and $x-8>0$), reject $x=-1$: $x=9$.
8. Inverse Functions
📘 Definition — Inverse function
A function is one-to-one if different inputs always give different outputs (every horizontal line meets the graph at most once — the horizontal line test). A one-to-one $f$ has an inverse function $f^{-1}$ with $$f^{-1}\big(f(x)\big)=x\quad\text{and}\quad f\big(f^{-1}(y)\big)=y.$$ The domain of $f^{-1}$ is the range of $f$ and vice versa; the graph of $f^{-1}$ is the reflection of the graph of $f$ in $y=x$. Note: $f^{-1}(x)\ne\dfrac1{f(x)}$.
✏️ Example 7 — Finding an inverse
Find the inverse of $f(x)=\dfrac{2x+1}{x-3}$.
💡 Show solution▶
Swap Write $y=\dfrac{2x+1}{x-3}$ and solve for $x$: $y(x-3)=2x+1\Rightarrow xy-2x=1+3y\Rightarrow x(y-2)=1+3y$.
Answer $x=\dfrac{3y+1}{y-2}$, so $f^{-1}(x)=\dfrac{3x+1}{x-2}$, $x\ne2$. (Check: $f^{-1}(f(0))=f^{-1}(-\tfrac13)=\dfrac{0}{-7/3}=0$ ✓.)
9. Trigonometry
In calculus, angles are measured in radians: an angle of $\theta$ radians cuts off an arc of length $\theta$ on the unit circle. Since the circumference is $2\pi$, $$360^\circ=2\pi\ \text{rad},\qquad 1^\circ=\frac{\pi}{180}\ \text{rad},\qquad 1\ \text{rad}=\frac{180^\circ}{\pi}\approx57.3^\circ.$$ The point on the unit circle at angle $\theta$ is $(\cos\theta,\sin\theta)$, and $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$, $\sec\theta=\dfrac1{\cos\theta}$, $\csc\theta=\dfrac1{\sin\theta}$, $\cot\theta=\dfrac{\cos\theta}{\sin\theta}$.
| $\theta$ | $0$ | $\frac\pi6$ | $\frac\pi4$ | $\frac\pi3$ | $\frac\pi2$ | $\pi$ | $\frac{3\pi}2$ |
|---|---|---|---|---|---|---|---|
| $\sin\theta$ | $0$ | $\frac12$ | $\frac{\sqrt2}2$ | $\frac{\sqrt3}2$ | $1$ | $0$ | $-1$ |
| $\cos\theta$ | $1$ | $\frac{\sqrt3}2$ | $\frac{\sqrt2}2$ | $\frac12$ | $0$ | $-1$ | $0$ |
| $\tan\theta$ | $0$ | $\frac{\sqrt3}3$ | $1$ | $\sqrt3$ | undefined | $0$ | undefined |
Memory aid: $\sin$ at $0,\frac\pi6,\frac\pi4,\frac\pi3,\frac\pi2$ is $\frac{\sqrt0}2,\frac{\sqrt1}2,\frac{\sqrt2}2,\frac{\sqrt3}2,\frac{\sqrt4}2$; cosine runs the same list backwards. Signs by quadrant: All positive (I), Sine (II), Tangent (III), Cosine (IV).
📐 Identities you will use constantly
$$\sin^2\theta+\cos^2\theta=1,\qquad 1+\tan^2\theta=\sec^2\theta,\qquad 1+\cot^2\theta=\csc^2\theta$$
$$\sin2\theta=2\sin\theta\cos\theta,\qquad \cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta=2\cos^2\theta-1$$
$$\sin(-\theta)=-\sin\theta\ \text{(odd)},\qquad \cos(-\theta)=\cos\theta\ \text{(even)},\qquad \sin\!\left(\tfrac\pi2-\theta\right)=\cos\theta$$
📘 Inverse trigonometric functions
To invert sine, cosine and tangent we restrict their domains so they become one-to-one:
- $\sin^{-1}x$ (arcsine): domain $[-1,1]$, range $\left[-\frac\pi2,\frac\pi2\right]$.
- $\cos^{-1}x$ (arccosine): domain $[-1,1]$, range $[0,\pi]$.
- $\tan^{-1}x$ (arctangent): domain $\mathbb R$, range $\left(-\frac\pi2,\frac\pi2\right)$.
✏️ Example 8 — Solving a trigonometric equation
Solve $2\sin x=1$ on $[0,2\pi)$.
💡 Show solution▶
Isolate $\sin x=\tfrac12$. The reference angle is $\sin^{-1}\tfrac12=\tfrac\pi6$.
Quadrants Sine is positive in quadrants I and II, so $x=\tfrac\pi6$ or $x=\pi-\tfrac\pi6=\tfrac{5\pi}6$.
Exercises
I
Find the domain (and the range where asked).
- 1 $f(x)=\sqrt{x-3}$ (domain and range)
- 2 $f(x)=\dfrac{1}{x^2-9}$
- 3 $f(x)=\ln(2x+4)$
- 4 $f(x)=\sqrt{9-x^2}$ (domain and range)
- 5 $f(x)=\dfrac{x+1}{x^2-x-6}$
💡 Show solutions▶
1 $x-3\ge0$: domain $[3,\infty)$; range $[0,\infty)$.
2 $x^2\ne9$: all real $x$ except $x=\pm3$.
3 $2x+4>0$: domain $(-2,\infty)$.
4 $9-x^2\ge0$: domain $[-3,3]$; the largest value is $3$ (at $x=0$), smallest $0$: range $[0,3]$ (upper semicircle).
5 $x^2-x-6=(x-3)(x+2)\ne0$: all real $x$ except $x=3$ and $x=-2$.
II
Let $f(x)=x^2+1$ and $g(x)=\sqrt x$.
- 1 $f\big(g(4)\big)$
- 2 $g\big(f(3)\big)$
- 3 $f(a+h)-f(a)$
- 4 $(f\circ g)(x)$ and its domain
- 5 Write $h(x)=(3x-1)^5$ as $F\big(G(x)\big)$.
💡 Show solutions▶
1 $g(4)=2$, $f(2)=5$.
2 $f(3)=10$, $g(10)=\sqrt{10}$.
3 $(a+h)^2+1-(a^2+1)=2ah+h^2$.
4 $f(\sqrt x)=x+1$, domain $[0,\infty)$ (inherited from $\sqrt x$).
5 $G(x)=3x-1$, $F(u)=u^5$.
III
Lines and quadratics.
- 1 Line through $(1,2)$ and $(3,8)$.
- 2 Line through $(-2,5)$ parallel to $y=4x+1$.
- 3 Solve $x^2-5x+6=0$.
- 4 Solve $2x^2+3x-2=0$.
- 5 Vertex of $y=x^2-4x+1$.
💡 Show solutions▶
1 $m=3$: $y=3x-1$.
2 $m=4$: $y-5=4(x+2)$, $y=4x+13$.
3 $(x-2)(x-3)=0$: $x=2,3$.
4 $(2x-1)(x+2)=0$: $x=\tfrac12,-2$.
5 $(2,-3)$.
IV
Factor or simplify.
- 1 $x^3-8$
- 2 $\dfrac{x^2-9}{x^2+x-6}$
- 3 $x^3+x^2-4x-4$
- 4 Rationalise $\dfrac{\sqrt{x+4}-2}{x}$
- 5 $\dfrac{\frac1x-\frac1{x+h}}{h}$
💡 Show solutions▶
1 $(x-2)(x^2+2x+4)$.
2 $\dfrac{x-3}{x-2}$, $x\ne-3$.
3 $(x+1)(x-2)(x+2)$.
4 $\dfrac{1}{\sqrt{x+4}+2}$.
5 $\dfrac{1}{x(x+h)}$.
V
Exponents and logarithms.
- 1 Simplify $\dfrac{\left(x^3y^{-2}\right)^2}{x^4y}$.
- 2 Evaluate $\log_232$.
- 3 Solve $e^{2x}=7$.
- 4 Expand $\ln\dfrac{x^2\sqrt{x+1}}{x-3}$.
- 5 Solve $\log_3x+\log_3(x-8)=2$.
💡 Show solutions▶
1 $\dfrac{x^6y^{-4}}{x^4y}=x^2y^{-5}=\dfrac{x^2}{y^5}$.
2 $2^5=32$, so $5$.
3 $x=\tfrac12\ln7$.
4 $2\ln x+\tfrac12\ln(x+1)-\ln(x-3)$.
5 $x=9$ ($x=-1$ rejected).
VI
Trigonometry.
- 1 Convert $150^\circ$ to radians and $\tfrac{7\pi}4$ to degrees.
- 2 Exact values: $\sin\tfrac{5\pi}6$, $\cos\tfrac{4\pi}3$, $\tan\tfrac{3\pi}4$.
- 3 Solve $2\sin x=1$ on $[0,2\pi)$.
- 4 Show that $\sec^2x-\tan^2x=1$.
- 5 Evaluate $\sin^{-1}\tfrac{\sqrt3}2$, $\tan^{-1}(-1)$, $\cos^{-1}\left(-\tfrac12\right)$.
💡 Show solutions▶
1 $150\cdot\tfrac{\pi}{180}=\tfrac{5\pi}6$; $\tfrac{7\pi}4\cdot\tfrac{180}{\pi}=315^\circ$.
2 $\tfrac12$, $-\tfrac12$, $-1$.
3 $\tfrac\pi6,\tfrac{5\pi}6$.
4 Divide $\sin^2x+\cos^2x=1$ by $\cos^2x$: $\tan^2x+1=\sec^2x$.
5 $\tfrac\pi3$, $-\tfrac\pi4$, $\tfrac{2\pi}3$.
VII
Inverse functions.
- 1 Find $f^{-1}$ for $f(x)=\dfrac{2x+1}{x-3}$.
- 2 Find $f^{-1}$ for $f(x)=e^{x-1}+2$ and state its domain.
- 3 Is $f(x)=x^2$ one-to-one on $\mathbb R$? If not, restrict it and find the inverse.
💡 Show solutions▶
1 $f^{-1}(x)=\dfrac{3x+1}{x-2}$.
2 $y-2=e^{x-1}\Rightarrow x=1+\ln(y-2)$: $f^{-1}(x)=1+\ln(x-2)$, domain $(2,\infty)$.
3 No: $f(-2)=f(2)$. On $[0,\infty)$ it is one-to-one with inverse $f^{-1}(x)=\sqrt x$.
VIII
Composition, inverses and symmetry.
- 1 Let $f(x)=2x-1$ and $g(x)=x^2$. Find $(f\circ g)(x)$ and $(g\circ f)(x)$, then evaluate both at $x=2$.
- 2 Let $f(x)=\sqrt x$ and $g(x)=x-5$. Find $f\circ g$ and $g\circ f$ together with their domains.
- 3 Let $f(x)=\dfrac1x$ and $g(x)=x+1$. Find $f\circ g$ and $g\circ f$ and the $x$-values each excludes.
- 4 Simplify $e^{\ln x}$ and $\ln\left(e^{x}\right)$, stating where each simplification is valid.
- 5 Show that $f(x)=3x+2$ and $g(x)=\dfrac{x-2}{3}$ are inverse functions.
- 6 Classify each as even, odd or neither: $f(x)=x^3-x$, $\;g(x)=x^2\cos x$, $\;h(x)=x^2+x$.
Composition machine — test parts 1–5 with numbers
💡 Show solutions▶
1 $(f\circ g)(x)=f(x^2)=2x^2-1$; $(g\circ f)(x)=g(2x-1)=(2x-1)^2=4x^2-4x+1$. At $x=2$: $2(4)-1=7$ and $3^2=9$ — composition is not commutative.
2 $(f\circ g)(x)=\sqrt{x-5}$ needs $x-5\ge0$: domain $[5,\infty)$. $(g\circ f)(x)=\sqrt x-5$ needs only $x\ge0$: domain $[0,\infty)$.
3 $(f\circ g)(x)=\dfrac{1}{x+1}$, all $x\ne-1$. $(g\circ f)(x)=\dfrac1x+1=\dfrac{x+1}{x}$, all $x\ne0$.
4 $e^{\ln x}=x$ for $x>0$ only (since $\ln x$ needs $x>0$); $\ln\left(e^x\right)=x$ for every real $x$ (since $e^x>0$ always).
5 $f(g(x))=3\cdot\dfrac{x-2}{3}+2=x$ and $g(f(x))=\dfrac{(3x+2)-2}{3}=x$. Both compositions give the identity, so $g=f^{-1}$.
6 $f(-x)=-x^3+x=-f(x)$: odd. $g(-x)=(-x)^2\cos(-x)=x^2\cos x=g(x)$: even. $h(-x)=x^2-x$, which is neither $h(x)$ nor $-h(x)$ (e.g. $h(1)=2$, $h(-1)=0$): neither.
IX
Exponential and logarithmic equations.
- 1 Solve $3^{2x-1}=27$.
- 2 Solve $5e^{0.2t}=40$ (exact and to two decimals).
- 3 Solve $\ln(x+2)-\ln x=\ln3$.
- 4 Solve $e^{2x}-5e^{x}+6=0$.
- 5 Write $2\ln x-3\ln y+\tfrac12\ln z$ as a single logarithm.
- 6 Evaluate $\log_48$ and $\log_{1/2}16$.
- 7 A culture starts with $500$ bacteria and doubles every $3$ hours, so $N(t)=500\cdot2^{t/3}$. When will there be $8000$ bacteria?
💡 Show solutions▶
1 $27=3^3$, so $2x-1=3$ and $x=2$.
2 $e^{0.2t}=8\Rightarrow0.2t=\ln8\Rightarrow t=5\ln8=15\ln2\approx10.40$.
3 $\ln\dfrac{x+2}{x}=\ln3\Rightarrow x+2=3x\Rightarrow x=1$. Check: $x=1>0$, so both logs are defined.
4 Put $u=e^x$: $u^2-5u+6=(u-2)(u-3)=0$, so $e^x=2$ or $e^x=3$: $x=\ln2$ or $x=\ln3$.
5 $\ln x^2-\ln y^3+\ln\sqrt z=\ln\dfrac{x^2\sqrt z}{y^3}$.
6 $4^{3/2}=8$, so $\log_48=\tfrac32$. $\left(\tfrac12\right)^{-4}=16$, so $\log_{1/2}16=-4$.
7 $500\cdot2^{t/3}=8000\Rightarrow2^{t/3}=16=2^4\Rightarrow t=12$ hours.
X
Trigonometric values, identities and equations.
- 1 Exact values: $\sin\tfrac{7\pi}6$, $\cos\left(-\tfrac\pi3\right)$, $\sec\tfrac{3\pi}4$, $\cot\tfrac{5\pi}3$.
- 2 If $\sin\theta=\tfrac35$ and $\theta$ is in quadrant II, find $\cos\theta$, $\tan\theta$ and $\sin2\theta$.
- 3 Solve $2\cos^2x-\cos x-1=0$ on $[0,2\pi)$.
- 4 Solve $\sin2x=\cos x$ on $[0,2\pi)$.
- 5 Show that $\left(1-\cos^2x\right)\left(1+\cot^2x\right)=1$.
- 6 A $6$ m ladder leans against a wall, making an angle of $\tfrac\pi3$ with the ground. How high up the wall does it reach?
- 7 Explain why $\sin^{-1}\left(\sin\tfrac{5\pi}6\right)=\tfrac\pi6$ and not $\tfrac{5\pi}6$.
💡 Show solutions▶
1 $\sin\tfrac{7\pi}6=-\tfrac12$ (quadrant III); $\cos\left(-\tfrac\pi3\right)=\cos\tfrac\pi3=\tfrac12$ (cosine is even); $\sec\tfrac{3\pi}4=\dfrac{1}{-\sqrt2/2}=-\sqrt2$; $\cot\tfrac{5\pi}3=\dfrac{1/2}{-\sqrt3/2}=-\tfrac{\sqrt3}3$.
2 $\cos^2\theta=1-\tfrac9{25}=\tfrac{16}{25}$ and cosine is negative in quadrant II: $\cos\theta=-\tfrac45$. Then $\tan\theta=-\tfrac34$ and $\sin2\theta=2\cdot\tfrac35\cdot\left(-\tfrac45\right)=-\tfrac{24}{25}$.
3 Factor: $(2\cos x+1)(\cos x-1)=0$. $\cos x=1\Rightarrow x=0$; $\cos x=-\tfrac12\Rightarrow x=\tfrac{2\pi}3,\tfrac{4\pi}3$.
4 $2\sin x\cos x-\cos x=\cos x\,(2\sin x-1)=0$. $\cos x=0\Rightarrow x=\tfrac\pi2,\tfrac{3\pi}2$; $\sin x=\tfrac12\Rightarrow x=\tfrac\pi6,\tfrac{5\pi}6$. (Do not divide by $\cos x$ — you would lose two solutions.)
5 $1-\cos^2x=\sin^2x$ and $1+\cot^2x=\csc^2x=\dfrac1{\sin^2x}$, so the product is $\sin^2x\cdot\dfrac1{\sin^2x}=1$ (for $\sin x\ne0$).
6 Height $=6\sin\tfrac\pi3=6\cdot\tfrac{\sqrt3}2=3\sqrt3\approx5.20$ m.
7 $\sin\tfrac{5\pi}6=\tfrac12$, and $\sin^{-1}$ must return an angle in $\left[-\tfrac\pi2,\tfrac\pi2\right]$ whose sine is $\tfrac12$; that angle is $\tfrac\pi6$. $\sin^{-1}(\sin\theta)=\theta$ only when $\theta$ already lies in that range.
🧪 Self-check quiz
0/6 correct
Q1. The domain of $\dfrac{1}{\sqrt{x-1}}$ is
Q2. The graph of $y=(x-2)^2+3$ is $y=x^2$ shifted
Q3. $(f\circ g)(x)$ means
Q4. $\ln\left(e^3\cdot e^{-5}\right)=$
Q5. $\cos\tfrac{2\pi}3=$
Q6. $\sin^{-1}x$ equals